Let $X_{i,t}$ be a random variable. Support $X_{i,t}$ converges to $X_{i,\infty}$ in distribution, whose the distribution is $f_{i,\infty}$. Let $Y_{t} = \max {\{f_{i,\infty} (X_{i,t}) : \forall i=1,\ldots,n\}}$ and $Y_{\infty} = \max {\{f_{i,\infty} (X_{i,\infty}) : \forall i=1,\ldots,n\}}$. Does $Y_{t}$ converges to $Y_{\infty}$ in distribution when $f_{i,\infty}$ is continuous for all $i=1,\ldots,n$?
The following is my proof.
As $f_{i,\infty}$ is continuous, $f_{i,\infty} (X_{i,t})$ converges to $f_{i,\infty} (X_{i,\infty})$ in distribution by continuous mapping theorem. In other words, $$\lim_{t \to \infty} \Pr (f_{i,\infty} (X_{i,t}) \leq a_{i} : \forall i=1,\ldots,n)=\Pr (f_{i,\infty} (X_{i,\infty}) \leq a_{i} : \forall i=1,\ldots,n)\tag{1}$$ for all $a_{1},\ldots,a_{n} \in [0,1]$. Then $$\begin{split}\Pr (Y_{t} \leq a) & = \lim_{t \to \infty} \Pr (f_{i,\infty} (X_{i,t}) \leq a : \forall i=1,\ldots,n)\\& = \Pr (f_{i,\infty} (X_{i,t}) \leq a : \forall i=1,\ldots,n)\\& = \Pr (Y_{\infty} \leq a)\end{split}$$ for all $a \in [0,1]$, by take $a_{1},\ldots,a_{n}=a$ in Equation (1). Thus $Y_{t}$ converges to $Y_{\infty}$ in distribuiton.
